Q.Minimise $Z = 13x - 15y$ subject to the constraints: $x + y \le 7$, $2x - 3y + 6 \ge 0$, $x \ge 0$, $y \ge 0$.

Concept: Linear Programming Graphical Method – we minimise $Z$ over the feasible region defined by the constraints.

Step 1 – Plot constraints

$x + y \le 7$ is a line through $(7,0)$ and $(0,7)$, shaded below.

$2x - 3y + 6 \ge 0$ rearranges to $y \le \frac{2x+6}{3}$, a line through $(-3,0)$ and $(0,2)$, shaded below it.

$x \ge 0$, $y \ge 0$ restrict to the first quadrant.

Step 2 – Find corner points

Intersection of $x+y=7$ and $2x-3y+6=0$: solving gives $x=3$, $y=4$.

Other corners: $(0,0)$, $(7,0)$, $(0,2)$.

Check $(0,0)$ satisfies all constraints. …

Concept. A linear objective over a bounded polygonal feasible region attains its optimum at a vertex, so we list the corner points and evaluate $Z$ at each.

Constraints: $x+y\le 7,\quad 2x-3y+6\ge 0\ (\text{i.e. } 2x-3y\ge-6),\quad x\ge 0,\ y\ge 0.$

Corner points of the feasible region:

• $(0,0)$
• $(7,0)$ - where $x+y=7$ meets the $x$-axis
• $(3,4)$ - where $x+y=7$ meets $2x-3y=-6$
• $(0,2)$ - where $2x-3y=-6$ meets the $y$-axis …

Method: Corner-Point Minimisation with a Negative Coefficient

Use this to minimise a linear objective such as $Z=ax-by$ (one coefficient negative) over a bounded region — the negative sign changes where the optimum lands but not the method.

Steps

Step 1: Plot the constraints and shade the feasible region.

Rewrite each inequality as an equation, draw the lines via intercepts, and use the origin test to keep the correct side. Watch a constraint written as $2x-3y+6\ge0$: rearrange to $2x-3y\ge-6$ before testing.

Step 2: Find the corner points, keeping only feasible intersections. …

Common Mistakes

Mistake 1: Sign errors when evaluating $Z=13x-15y$.

Why it's wrong: the $-15y$ term makes larger $y$ decrease $Z$; mishandling the sign flips which corner looks smallest. At $(0,2)$, $Z=0-30=-30$ (the minimum), while $(7,0)$ gives $+91$. Correct approach: substitute carefully with the negative coefficient and compare signed values.

Mistake 2: Treating $(0,7)$ as a corner. …